CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 14 Polygons Ex 14B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A
• RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14B

Question 1.
Solution:
In a pentagon, no. of diagonals
\(=\frac { n\left( n-3 \right) }{ 2 }\)
\(=\frac { 5\left( 5-3 \right) }{ 2 }\)
\( =\frac { 5\times 2 }{ 2 } \)
= 5 (a)

Question 2.
Solution:
In a hexagon, no. of diagonals
\(=\frac { n\left( n-3 \right) }{ 2 }\)
\(=\frac { 6\left( 6-3 \right) }{ 2 }\)
\( =\frac { 6\times 3 }{ 2 } \)
= 9 (c)

Question 3.
Solution:
In an octagon, no. of diagonals
\(=\frac { n\left( n-3 \right) }{ 2 }\)
\(=\frac { 8\left( 8-3 \right) }{ 2 }\)
\( =\frac { 8\times 5 }{ 2 } \)
= 20 (d)

Question 4.
Solution:
In a polygon of 12 sides, no. of diagonals
\(=\frac { n\left( n-3 \right) }{ 2 }\)
\(=\frac { 12\left( 12-3 \right) }{ 2 }\)
\( =\frac { 12\times 9 }{ 2 } \)
= 54 (c)

Question 5.
Solution:
A polygon has 27 diagonal

Either n – 9 = 0, then n = 9
or n + 6 = 0, then n = – 6 but it is not possible being negative
No. of sides = 9 (c)

Question 6.
Solution:
Angles of a pentagon are x°, (x + 20)°, (x + 40)°, (x + 60°) and (x + 80)°
But sum of angle of a pentagon

Question 7.
Solution:
Measure of each exterior angle = 40°
No. of sides = \(\frac { { 360 }^{ o } }{ 40 }\)9 sides (b)

Question 8.
Solution:
Each interior angle of a polygon = 108°

Question 9.
Solution:
Each interior angle = 135°

Question 10.
Solution:
Let each exterior angle = x, then
Each interior angles = 3n
But sum of angle = 180°
x + 3x = 180°
=>4x = 180°
=> x = 45°
No. of sides = \(\frac { { 360 }^{ o } }{ 45 } \)
= 8 sides (b)

Question 11.
Solution:
Each interior angles of decagon

Question 12.
Solution:
Sum of all interior angles of a hexagon
= (2n – 4) x right angle
= (2 x 6 – 4) right angle
= 8 right angles (b)

Question 13.
Solution:
Sum of all interior angles of polygon = 1080°
Let n be the number of sides, then
(2n – 4) x 90°= 1080°

Question 14.
Solution:
Difference between each interior and exterior angle = 108°
Then each interior angle = x + 108°
x + x + 108°= 180°
(Sum of both angles = 180°)
=> 2x = 180° – 108° = 72°
x = \(\\ \frac { 72 }{ 2 } \)
= 36°
No. of sides = \( \frac { { 360 }^{ o } }{ { 36 }^{ o } } \)
= 10 (d)

Hope given RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14B are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 14 Polygons Ex 14A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A
• RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14B

Question 1.
Solution:
We know that sum of exterior angles of a polygon is 360°
Then,
(i) Pentagon’s exterior angle = \(\frac { { 360 }^{ o } }{ 5 } \)
= 72°

Question 2.
Solution:
Each exterior angle a n sided polygon = 50°
No of sides = \(\frac { { 360 }^{ o } }{ 50 } \)
= \(7\frac { 1 }{ 5 } \)
Which is not possible to have \(7\frac { 1 }{ 5 } \) sides
Which is not a whole number

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Question 3.
Solution:
We know that each interior angle of a regular polygon of n sides = \(\\ \frac { 2n-4 }{ n } \) right angle
(i) Polygon having 10 sides, each interior

Question 4.
Solution:
Let interior angle of a polygon having n sides = 100°
\(\\ \frac { 2n-4 }{ n } \) x 90° = 100°
=>\(\\ \frac { 2n-4 }{ n } \)
= \(\\ \frac { 100 }{ 90 } \)
= \(\\ \frac { 10 }{ 9 } \)

Question 5.
Solution:
We know that sum of all interior angles = 2n – 4 right angles
(i) Pentagon
Sum of its angles = (2 x 5 – 4) x 90°
= 6 x 90° = 540°

Question 6.
Solution:
We know that number of diagonal of polygon having n sides = \(\frac { n\left( n-3 \right) }{ 2 } \)
(i) In heptagon, no of diagonals = \(\frac { 7\left( 7-3 \right) }{ 2 } \)

Question 7.
Solution:
We know that each exterior angle
= \( \frac { { 360 }^{ o } }{ n } \)
Where n sides are of polygon
(i) Each exterior angle = 40°

Question 8.
Solution:
We know that sum of all exterior angle of a polygon – 360°
Exterior ∠A + ∠B + ∠C + ∠D = 360°
=> 115° + x + 90° + 50° = 360°
=> 255° + x + 360°
=> x = 360° – 255°
=> x = 105°

Question 9.
Solution:
In the given figure polygon is of 5 sides and each interior angle is x
\(x=\frac { 2x-4 }{ n } \times { 90 }^{ o }=\frac { 2\times 5-4 }{ 5 } \times { 90 }^{ o } \)
= \(=\frac { 6 }{ 5 } \times { 90 }^{ o } \)
= 108°

Hope given RS Aggarwal Solutions Class 8 Chapter 14 Polygons Ex 14A are helpful to complete your math homework.

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RS Aggarwal Class 8 Solutions Chapter 13 Time and Work Ex 13B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13A
• RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13B

Question 1.
Solution:
A’s 1 day’s work = \(\\ \frac { 1 }{ 10 } \)
B’s 1 day’s work = \(\\ \frac { 1 }{ 15 } \)

Question 2.
Solution:
A man’s 1 day work = \(\\ \frac { 1 }{ 5 } \)
Man and his son’s 1 days work = \(\\ \frac { 1 }{ 3 } \)

Question 3.
Solution:
A’s 1 day’s work = \(\\ \frac { 1 }{ 16 } \)
B’s 1 day’s work = \(\\ \frac { 1 }{ 12 } \)

Question 4.
Solution:
Let B can do a work in = x days
Then A can do the work

Question 5.
Solution:
Let B’s 1 day’s work = x
Then A’s 1 day’s work = 2x

Question 6.
Solution:
Total wages = Rs. 3000
A’s 1 day’s work = \(\\ \frac { 1 }{ 10 } \)
B’s 1 day’s work = \(\\ \frac { 1 }{ 15 } \)

Question 7.
Solution:
Ratio in the rates of working of A and B = 3:4
Ratio in time = \(\\ \frac { 1 }{ 3 } \) : \(\\ \frac { 1 }{ 4 } \)
= \(\\ \frac { 4:3 }{ 12 } \)
= 4 : 3 (c)

Question 8.
Solution:
A and B’s 1 day’s wok = \(\\ \frac { 1 }{ 12 } \)
B and C’s 1 day’s work = \(\\ \frac { 1 }{ 20 } \)
C and A’s 1 day’s work = \(\\ \frac { 1 }{ 15 } \)

Question 9.
Solution:
3 men = 5 women
1 man = \(\\ \frac { 5 }{ 3 } \) women 5
6 men = \(\\ \frac { 5 }{ 3 } \) x 6 = 10 women

Question 10.
Solution:
A’s 1 day’s work = \(\\ \frac { 1 }{ 15 } \)
Then B’s 1 day’s work = \(\\ \frac { 1 }{ 10 } \) x \(\\ \frac { 100+50 }{ 100 } \)
= \(\\ \frac { 1 }{ 15 } \) x \(\\ \frac { 150 }{ 100 } \)
= \(\\ \frac { 1 }{ 10 } \)
B will finish the work in = 10 days (a)

Question 11.
Solution:
A’s 1 hour’s work = \(\\ \frac { 2 }{ 15 } \)
A and B’s ratio in work = \(\\ \frac { 100-20 }{ 100 } \) : 1
= \(\\ \frac { 80 }{ 100 } \) : 1
= \(\\ \frac { 4 }{ 5 } \) : 1

Question 12.
Solution:
A’s 1 day’s work = \(\\ \frac { 1 }{ 20 } \)
B’s 1 day’s work = \(\\ \frac { 1 }{ 12 } \)

Question 13.
Solution:
A’s 1 days work = \(\\ \frac { 1 }{ 25 } \)
B’s 1 days work = \(\\ \frac { 1 }{ 20 } \)

Question 14.
Solution:
First pipe 1 minutes work = \(\\ \frac { 1 }{ 20 } \)
Second pipe 1 minutes work = \(\\ \frac { 1 }{ 30 } \)

Question 15.
Solution:
First tap’s 1 hours work to fill = \(\\ \frac { 1 }{ 8 } \)
Second tap’s 1 hours work to empty = \(\\ \frac { 1 }{ 16 } \)
Both 1 hour can fill the cistern

Question 16.
Solution:
First pump’s 1 hr work to fill = \(\\ \frac { 1 }{ 2 } \)
Due to leakage, tank is filled in \(2\frac { 1 }{ 3 } \) hour

Question 17.
Solution:
Answer = (b)
First inlet pipe’s 1 hour work = \(\\ \frac { 1 }{ 10 } \)
Second inlet pipe’s 1 hour work = \(\\ \frac { 1 }{ 12 } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 13 Time and Work Ex 13A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13A
• RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13B

Question 1.
Solution:
Rajan’s one day’s work = \(\\ \frac { 1 }{ 24 } \)
Amit’s one day’s work = \(\\ \frac { 1 }{ 30 } \)

Question 2.
Solution:
Ravi’s one hours = \(\\ \frac { 1 }{ 15 } \)
Both’s one day’s work = \(\\ \frac { 1 }{ 12 } \)

or 6 hours, 40 minutes.

Question 3.
Solution:
A and B both’s one day’s work = \(\\ \frac { 1 }{ 6 } \)
A’s alone’s one day’s work = \(\\ \frac { 1 }{ 9 } \)

Question 4.
Solution:
Raju and Siraj’s 1 hour work = \(\\ \frac { 1 }{ 6 } \)
Raju’s alone 1 hour work = \(\\ \frac { 1 }{ 15 } \)

Question 5.
Solution:
A’s one day’s work = \(\\ \frac { 1}{ 10 } \)
B’s one day’s work = \(\\ \frac { 1 }{ 12 } \)

Question 6.
Solution:
A’s 1 hour work = \(\\ \frac { 1 }{ 24 } \)
B’s 1 hour work = \(\\ \frac { 1 }{ 16 } \)

Question 7.
Solution:
A,B and C’s 1 hr work = \(\\ \frac { 1 }{ 8 } \)
A’s 1 hour work = \(\\ \frac { 1 }{ 20 } \)
B’s 1 hour work = \(\\ \frac { 1 }{ 24 } \)

Question 8.
Solution:
A’s one day’s work = \(\\ \frac { 1 }{ 16 } \)
B’s one days work = \(\\ \frac { 1 }{ 12 } \)

Question 9.
Solution:
A’s 1 day’s work = \(\\ \frac { 1 }{ 14 } \)
B’s 1 day’s work = \(\\ \frac { 1 }{ 21 } \)

Question 10.
Solution:
A can do \(\\ \frac { 2 }{ 3 } \) work in = 16 days
A’s 1 days work = \(\\ \frac { 2 }{ 3 } \) x \(\\ \frac { 1 }{ 16 } \) = \(\\ \frac { 1 }{ 24 } \)

Question 11.
Solution:
A’s one day’s work = \(\\ \frac { 1 }{ 15 } \)
B’s one day’s work = \(\\ \frac { 1 }{ 12 } \)

Question 12.
Solution:
A and B’s one day’s work = \(\\ \frac { 1 }{ 18 } \)
B and C’s one day’s work = \(\\ \frac { 1 }{ 24 } \)

A, B and C’s one days work = \(\frac { 1 }{ 2\times 2 } \)
= \(\\ \frac { 1 }{ 16 } \)
A, B and C can do the work in 16 days.

Question 13.
Solution:
A and B’s one days work = \(\\ \frac { 1 }{ 12 } \)
B and C’s one day’s work = \(\\ \frac { 1 }{ 15 } \)

Question 14.
Solution:
A’s one hr work =\(\\ \frac { 1 }{ 10 } \)
B’s one hr work = \(\\ \frac { 1 }{ 15 } \)

Question 15.
Solution:
Pipe A’s one hour’s work for filling the tank = \(\\ \frac { 1 }{ 5 } \)
Pipe B’s one hour’s work for emptying = \(\\ \frac { 1 }{ 6 } \)

Question 16.
Solution:
Tap A’s one hour’s work = \(\\ \frac { 1 }{ 6 } \)
Tap B’s one hour’s work = \(\\ \frac { 1 }{ 8 } \)
Tap C’s one hour’s work = \(\\ \frac { 1 }{ 12 } \)
A, B and C’s together one hour’s work

Question 17.
Solution:
Inlet A’s 1 minutes work = \(\\ \frac { 1 }{ 12 } \)
Inlet B’s 1 minutes work = \(\\ \frac { 1 }{ 15 } \)

Question 18.
Solution:
The inlet pipe’s 1 hour’s work = \(\\ \frac { 1 }{ 9 } \)
The leak and inlet’s 1 hours work = \(\\ \frac { 1 }{ 10 } \)
Leak’s 1 hour work = \(\frac { 1 }{ 9 } -\frac { 1 }{ 10 } \)
= \(\\ \frac { 10-9 }{ 90 } \)
= \(\\ \frac { 1 }{ 90 } \)
The leak can empty the cistern in = 90 hours Ans.

Question 19.
Solution:
Inlet pipe A’s one hour’s work = \(\\ \frac { 1 }{ 6 } \)
Inlet pipe B’s one hour’s work = \(\\ \frac { 1 }{ 8 } \)

Hope given RS Aggarwal Solutions Class 8 Chapter 13 Time and Work Ex 13A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 12 Direct and Inverse Proportions Ex 12C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12A
• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12B
• RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12C

OBJECTIVE QUESTIONS :
Tick the correct answer in each of the following :

Question 1.
Solution:
Answer = (d)
Cost of 14 kg of pulses = Rs 882
Cost of 1 kg of pulses Rs \(\\ \frac { 882 }{ 14 } \)
Cost of 22 kg of pulses = Rs \(\frac { 882\times 22 }{ 14 } \)
= 63 x 22
= Rs 1386

Question 2.
Solution:
Let x be oranges which can be bought for Rs. 33.80
8 : x : : 10.40 : 33.80
=> x × 10.40 = 8 x 33.80
=> \(\frac { 8\times 33.80 }{ 10.40 } \)
=> \(\frac { 8\times 3380 }{ 1040 } \)
= 26
∴ No. of oranges = 26 Ans. (c)

Question 3.
Solution:
No. of bottles 420 x
Time 3 5
More time, more bottles
By direct proportion
420 : x :: 3 : 5
x = \(\frac { 420\times 5 }{ 3 } \)
= 700
∴ No. of bottle will be 700 (b)

Question 4.
Solution:
Distance covered 75 km x
Time taken 60 min. 20 min.
Less time, less distance By direct proportion,
75 : x :: 60 : 20
x = \(\frac { 75\times 20 }{ 60 } \)
= 25
∴ Distance covered = 25 km (a)

Question 5.
Solution:
No. of sheets 12 : x
Weight 40 g : 1000 g
More weight, more sheets
By direct proportion 12 : x :: 40 : 1000
x = \(\frac { 12\times 1000 }{ 40 } \)
= 300
∴ No. of sheets = 300 (c)

Question 6.
Solution:
Let x be the height of tree
Height of pole 14 m : x m
Length of shadow 10 m : 7 m
Less shadow, less height
By direct proportion 14 : x :: 10 : 7
x = \(\frac { 14\times 7 }{ 10 } \)
= \(\\ \frac { 98 }{ 10 } \)
= 9.8 m
∴ Height of the = 9.8 m (b)

Question 7.
Solution:
Let actual length of bacteria = x cm
Enlarged (times) 50000
Length 5 cm
Then actual length (x)
Then \(\frac { x\times 50000 }{ -4 } \)
= 5
=> x = \(\\ \frac { 5 }{ 50000 } \)
= \(\\ \frac { 1 }{ 10000 } \)
= 10 cm (c)

Question 8.
Solution:
No. of pipes 6 : 5
Time taken to 120 min : x min
fill the tank
Less pipes, more time
By inverse proportion 6 : 5 :: x : 120
x = \(\frac { 6\times 120 }{ 5 } \)
= 144 (b)
∴ Time taken = 144 minutes

Question 9.
Solution:
Let number of days = x, then
Persons 3 : 4
(Time taken to build a wall) 4 : x
More person, less time take
By inverse proportion,
3 : 4 :: x : 4
x = \(\frac { 4\times 3 }{ 4 } \)
= 3 (b)
∴ Time taken to build the wall = 3 days

Question 10.
Solution:
Let time taken will be x hrs
Speed 60 km/h : 80 km/h
Time taken to 2hr : x
reach
(More speed, less time)
By inverse proportion 60 : 80 :: x : 2
x = \(\frac { 60\times 2 }{ 80 } \)
= \(\\ \frac { 3 }{ 2 } \)
∴ Time take \(\\ \frac { 3 }{ 2 } \) hours or 1 hr. 30 m in. (a)

Hope given RS Aggarwal Solutions Class 8 Chapter 12 Direct and Inverse Proportions Ex 12C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.